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Class 12 › Physics › Moving Charges and Magnetism
CBSE2023Class Class 12 · Physics1 Marks · Long✅ Verified
Case Study (Charged particles): Particle 1 (mass $m$, charge $-q$) and particle 2 (mass $\frac{m}{2}$, charge $+2q$) in a magnetic field $B$. Q (OR part a): If the period of revolution for particle 1 is 4 s, then for particle 2 the period will be:
✅ Answer & Solution
Time period of circular motion in a magnetic field: $$T = \frac{2\pi m}{qB}$$ $T \propto \dfrac{m}{q}$. For particle 1: $T_1 \propto \dfrac{m}{q}$. For particle 2: $T_2 \propto \dfrac{m/2}{2q} = \dfrac{m}{4q} = \dfrac14\cdot\dfrac{m}{q}$. $$\frac{T_2}{T_1} = \frac{1}{4} \Rightarrow T_2 = \frac{T_1}{4} = \frac{4}{4} = 1\ \text{s}$$ Hence the correct option is (A).
